Problem: Santa has an infinite number of candies for each of m flavours. You are given a rooted tree with n vertices. The root of the tree is the vertex 1. Each vertex contains exactly one candy. The i-th vertex has a candy of flavour f_i.

Sometimes Santa fears that candies of flavour k have melted. He chooses any vertex x randomly and sends the subtree of x to the Bakers for a replacement. In a replacement, all the candies with flavour k are replaced with a new candy of the same flavour. The candies which are not of flavour k are left unchanged. After the replacement, the tree is restored.

The actual cost of replacing one candy of flavour k is c_k (given for each k). The Baker keeps the price fixed in order to make calculation simple. Every time when a subtree comes for a replacement, the Baker charges C, no matter which subtree it is and which flavour it is.

Suppose that for a given flavour k the probability that Santa chooses a vertex for replacement is same for all the vertices. You need to find out the expected value of error in calculating the cost of replacement of flavour k. The error in calculating the cost is defined as follows.

$$$ Error\ E(k) =\ (Actual Cost\ –\ Price\ charged\ by\ the\ Bakers) ^ 2.$$$

Note that the actual cost is the cost of replacement of one candy of the flavour k multiplied by the number of candies in the subtree.

Also, sometimes Santa may wish to replace a candy at vertex x with a candy of some flavour from his pocket.

You need to handle two types of operations: 

  * Change the flavour of the candy at vertex x to w. 
  * Calculate the expected value of error in calculating the cost of replacement for a given flavour k. 

Input

The first line of the input contains four integers n (2 ⩽ n ⩽ 5 ⋅ 10^4), m, q, C (1 ⩽ m, q ⩽ 5 ⋅ 10^4, 0 ⩽ C ⩽ 10^6) — the number of nodes, total number of different flavours of candies, the number of queries and the price charged by the Bakers for replacement, respectively.

The second line contains n integers f_1, f_2, ..., f_n (1 ⩽ f_i ⩽ m), where f_i is the initial flavour of the candy in the i-th node.

The third line contains n - 1 integers p_2, p_3, ..., p_n (1 ⩽ p_i ⩽ n), where p_i is the parent of the i-th node.

The next line contains m integers c_1, c_2, ... c_m (1 ⩽ c_i ⩽ 10^2), where c_i is the cost of replacing one candy of flavour i.

The next q lines describe the queries. Each line starts with an integer t (1 ⩽ t ⩽ 2) — the type of the query.

If t = 1, then the line describes a query of the first type. Two integers x and w follow (1 ⩽ x ⩽ n, 1 ⩽ w ⩽ m), it means that Santa replaces the candy at vertex x with flavour w.

Otherwise, if t = 2, the line describes a query of the second type and an integer k (1 ⩽ k ⩽ m) follows, it means that you should print the expected value of the error in calculating the cost of replacement for a given flavour k.

The vertices are indexed from 1 to n. Vertex 1 is the root.

Output

Output the answer to each query of the second type in a separate line.

Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. The checker program considers your answer correct if and only if (|a-b|)/(max(1,b))⩽ 10^{-6}.

Example

Input

3 5 5 7
3 1 4
1 1
73 1 48 85 89
2 1
2 3
1 2 3
2 1
2 3


Output

2920.333333333333
593.000000000000
49.000000000000
3217.000000000000

Note

For 1-st query, the error in calculating the cost of replacement for flavour 1 if vertex 1, 2 or 3 is chosen are 66^2, 66^2 and (-7)^2 respectively. Since the probability of choosing any vertex is same, therefore the expected value of error is (66^2+66^2+(-7)^2)/(3).

Similarly, for 2-nd query the expected value of error is (41^2+(-7)^2+(-7)^2)/(3).

After 3-rd query, the flavour at vertex 2 changes from 1 to 3.

For 4-th query, the expected value of error is ((-7)^2+(-7)^2+(-7)^2)/(3).

Similarly, for 5-th query, the expected value of error is (89^2+41^2+(-7)^2)/(3).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T, class F = function<T(const T&, const T&)>>
class SparseTable {
 public:
  int n;
  vector<vector<T>> mat;
  F func;
  SparseTable(){};
  SparseTable(const vector<T>& a, const F& f) : func(f) {
    n = static_cast<int>(a.size());
    int max_log = 32 - __builtin_clz(n);
    mat.resize(max_log);
    mat[0] = a;
    for (int j = 1; j < max_log; j++) {
      mat[j].resize(n - (1 << j) + 1);
      for (int i = 0; i <= n - (1 << j); i++) {
        mat[j][i] = func(mat[j - 1][i], mat[j - 1][i + (1 << (j - 1))]);
      }
    }
  }
  T get(int from, int to) const {
    assert(0 <= from && from <= to && to <= n - 1);
    int lg = 32 - __builtin_clz(to - from + 1) - 1;
    return func(mat[lg][from], mat[lg][to - (1 << lg) + 1]);
  }
};
class tree {
 public:
  ;
  int n;
  vector<vector<int>> gr;
  vector<int> in, out, tour, dep, par, sub;
  SparseTable<int> st;
  tree(){};
  tree(int _n) {
    n = _n;
    gr.resize(n);
    in.resize(n);
    out.resize(n);
    dep.resize(n);
    par.resize(n);
    sub.resize(n, 1);
  }
  void add(int u, int v) {
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  void dfs(int v, int p) {
    in[v] = tour.size();
    tour.push_back(v);
    if (~p) dep[v] = dep[p] + 1;
    par[v] = p;
    for (int i : gr[v]) {
      if (i == p) continue;
      sub[v] += sub[i];
      dfs(i, v);
      tour.push_back(v);
    }
    out[v] = tour.size() - 1;
  }
  void prep() {
    dfs(0, -1);
    st = SparseTable<int>(tour, [&](int x, int y) {
      return (dep[x] == dep[y] ? min(x, y) : (dep[x] > dep[y] ? y : x));
    });
  }
  int parent(int u, int v) {
    return st.get(min(in[u], in[v]), max(out[u], out[v]));
  }
  pair<map<int, int>, vector<int>> to_tree(vector<int> id) {
    sort(id.begin(), id.end(), [&](int x, int y) { return in[x] < in[y]; });
    int sz = id.size();
    for (int i = 0; i < sz - 1; i++) {
      id.push_back(parent(id[i], id[i + 1]));
    }
    sort(id.begin(), id.end(), [&](int x, int y) { return in[x] < in[y]; });
    id.resize(unique(id.begin(), id.end()) - id.begin());
    sz = id.size();
    vector<int> p(sz);
    map<int, int> to;
    for (int i = 0; i < sz; i++) to[id[i]] = i;
    vector<int> que;
    for (int i = 0; i < sz; i++) {
      while (!que.empty() && out[id[que.back()]] < in[id[i]]) que.pop_back();
      if (que.empty())
        p[i] = -1;
      else
        p[i] = que.back();
      que.push_back(i);
    }
    return make_pair(to, p);
  }
};
class node {
 public:
  int id;
  node* l;
  node* r;
  node* p;
  bool rev;
  int sz;
  long long sum;
  long long dep;
  long long total_sum;
  long long total_dep;
  long long ad;
  node(int _id) {
    id = _id;
    l = r = p = NULL;
    rev = false;
    sz = 1;
    sum = 0;
    dep = 0;
    total_sum = 0;
    total_dep = 0;
    ad = 0;
  }
  void unsafe_reverse() {
    rev ^= 1;
    swap(l, r);
    pull();
  }
  void unsafe_apply(long long v) {
    total_sum += v * total_dep;
    sum += v * dep;
    ad += v;
  }
  void push() {
    if (rev) {
      if (l != NULL) {
        l->unsafe_reverse();
      }
      if (r != NULL) {
        r->unsafe_reverse();
      }
      rev = 0;
    }
    if (ad) {
      if (l != NULL) {
        l->unsafe_apply(ad);
      }
      if (r != NULL) {
        r->unsafe_apply(ad);
      }
      ad = 0;
    }
  }
  void pull() {
    sz = 1;
    total_sum = sum;
    total_dep = dep;
    if (l != NULL) {
      l->p = this;
      sz += l->sz;
      total_sum += l->total_sum;
      total_dep += l->total_dep;
    }
    if (r != NULL) {
      r->p = this;
      sz += r->sz;
      total_sum += r->total_sum;
      total_dep += r->total_dep;
    }
  }
};
namespace splay_tree {
bool is_bst_root(node* v) {
  if (v == NULL) {
    return false;
  }
  return (v->p == NULL || (v->p->l != v && v->p->r != v));
}
void rotate(node* v) {
  node* u = v->p;
  assert(u != NULL);
  u->push();
  v->push();
  v->p = u->p;
  if (v->p != NULL) {
    if (v->p->l == u) {
      v->p->l = v;
    }
    if (v->p->r == u) {
      v->p->r = v;
    }
  }
  if (v == u->l) {
    u->l = v->r;
    v->r = u;
  } else {
    u->r = v->l;
    v->l = u;
  }
  u->pull();
  v->pull();
}
void splay(node* v) {
  if (v == NULL) {
    return;
  }
  while (!is_bst_root(v)) {
    node* u = v->p;
    if (!is_bst_root(u)) {
      if ((u->l == v) ^ (u->p->l == u)) {
        rotate(v);
      } else {
        rotate(u);
      }
    }
    rotate(v);
  }
}
pair<node*, int> find(node* v, const function<int(node*)>& go_to) {
  if (v == NULL) {
    return {NULL, 0};
  }
  splay(v);
  int dir;
  while (true) {
    v->push();
    dir = go_to(v);
    if (dir == 0) {
      break;
    }
    node* u = (dir == -1 ? v->l : v->r);
    if (u == NULL) {
      break;
    }
    v = u;
  }
  splay(v);
  return {v, dir};
}
node* get_leftmost(node* v) {
  return find(v, [&](node*) { return -1; }).first;
}
node* get_rightmost(node* v) {
  return find(v, [&](node*) { return 1; }).first;
}
node* get_kth(node* v, int k) {
  pair<node*, int> p = find(v, [&](node* u) {
    if (u->l != NULL) {
      if (u->l->sz > k) {
        return -1;
      }
      k -= u->l->sz;
    }
    if (k == 0) {
      return 0;
    }
    k--;
    return 1;
  });
  return (p.second == 0 ? p.first : NULL);
}
int get_position(node* v) {
  splay(v);
  return (v->l != NULL ? v->l->sz : 0);
}
node* get_bst_root(node* v) {
  splay(v);
  return v;
}
pair<node*, node*> split(node* v, const function<bool(node*)>& is_right) {
  if (v == NULL) {
    return {NULL, NULL};
  }
  pair<node*, int> p = find(v, [&](node* u) { return is_right(u) ? -1 : 1; });
  v = p.first;
  v->push();
  if (p.second == -1) {
    node* u = v->l;
    if (u == NULL) {
      return {NULL, v};
    }
    v->l = NULL;
    u->p = v->p;
    u = get_rightmost(u);
    v->p = u;
    v->pull();
    return {u, v};
  } else {
    node* u = v->r;
    if (u == NULL) {
      return {v, NULL};
    }
    v->r = NULL;
    v->pull();
    return {v, u};
  }
}
pair<node*, node*> split_leftmost_k(node* v, int k) {
  return split(v, [&](node* u) {
    int left_and_me = (u->l != NULL ? u->l->sz : 0) + 1;
    if (k >= left_and_me) {
      k -= left_and_me;
      return false;
    }
    return true;
  });
}
node* merge(node* v, node* u) {
  if (v == NULL) {
    return u;
  }
  if (u == NULL) {
    return v;
  }
  v = get_rightmost(v);
  assert(v->r == NULL);
  splay(u);
  v->push();
  v->r = u;
  v->pull();
  return v;
}
int count_left(node* v, const function<bool(node*)>& is_right) {
  if (v == NULL) {
    return 0;
  }
  pair<node*, int> p = find(v, [&](node* u) { return is_right(u) ? -1 : 1; });
  node* u = p.first;
  return (u->l != NULL ? u->l->sz : 0) + (p.second == 1);
}
node* add(node* r, node* v, const function<bool(node*)>& go_left) {
  pair<node*, node*> p = split(r, go_left);
  return merge(p.first, merge(v, p.second));
}
node* remove(node* v) {
  splay(v);
  v->push();
  node* x = v->l;
  node* y = v->r;
  v->l = v->r = NULL;
  node* z = merge(x, y);
  if (z != NULL) {
    z->p = v->p;
  }
  v->p = NULL;
  v->push();
  v->pull();
  return z;
}
node* next(node* v) {
  splay(v);
  v->push();
  if (v->r == NULL) {
    return NULL;
  }
  v = v->r;
  while (v->l != NULL) {
    v->push();
    v = v->l;
  }
  splay(v);
  return v;
}
node* prev(node* v) {
  splay(v);
  v->push();
  if (v->l == NULL) {
    return NULL;
  }
  v = v->l;
  while (v->r != NULL) {
    v->push();
    v = v->r;
  }
  splay(v);
  return v;
}
int get_size(node* v) {
  splay(v);
  return (v != NULL ? v->sz : 0);
}
template <typename... T>
void apply(node* v, T... args) {
  splay(v);
  v->unsafe_apply(args...);
}
void reverse(node* v) {
  splay(v);
  v->unsafe_reverse();
}
}  // namespace splay_tree
using namespace splay_tree;
template <bool rooted>
class link_cut_tree {
 public:
  int n;
  vector<node*> nodes;
  link_cut_tree(){};
  link_cut_tree(int _n) : n(_n) {
    nodes.resize(n);
    for (int i = 0; i < n; i++) {
      nodes[i] = new node(i);
    }
  }
  int add_node() {
    int id = (int)nodes.size();
    nodes.push_back(new node(id));
    return id;
  }
  void expose(node* v) {
    node* r = NULL;
    node* u = v;
    while (u != NULL) {
      splay(u);
      u->push();
      u->r = r;
      u->pull();
      r = u;
      u = u->p;
    }
    splay(v);
    assert(v->p == NULL);
  }
  int get_root(int i) {
    node* v = nodes[i];
    expose(v);
    return get_leftmost(v)->id;
  }
  bool link(int i, int j) {
    if (i == j) {
      return false;
    }
    node* v = nodes[i];
    node* u = nodes[j];
    if (rooted) {
      splay(v);
      assert(v->p == NULL && v->l == NULL);
    } else {
      make_root(i);
    }
    expose(u);
    if (v->p != NULL) {
      return false;
    }
    v->p = u;
    return true;
  }
  bool cut(int i, int j) {
    if (i == j) {
      return false;
    }
    node* v = nodes[i];
    node* u = nodes[j];
    expose(u);
    splay(v);
    if (v->p != u) {
      if (rooted) {
        return false;
      }
      swap(u, v);
      expose(u);
      splay(v);
      if (v->p != u) {
        return false;
      }
    }
    v->p = NULL;
    return true;
  }
  bool connected(int i, int j) {
    if (i == j) {
      return true;
    }
    node* v = nodes[i];
    node* u = nodes[j];
    expose(v);
    assert(v->p == NULL);
    expose(u);
    return v->p != NULL;
  }
  int lca(int i, int j) {
    if (i == j) {
      return i;
    }
    node* v = nodes[i];
    node* u = nodes[j];
    expose(v);
    assert(v->p == NULL);
    expose(u);
    if (v->p == NULL) {
      return -1;
    }
    splay(v);
    if (v->p == NULL) {
      return v->id;
    }
    return v->p->id;
  }
  bool is_ancestor(int i, int j) {
    if (i == j) {
      return true;
    }
    node* v = nodes[i];
    node* u = nodes[j];
    expose(u);
    assert(u->p == NULL);
    splay(v);
    return v->p == NULL && u->p != NULL;
  }
  void make_root(int i) {
    assert(!rooted);
    node* v = nodes[i];
    expose(v);
    assert(v->r == NULL);
    reverse(v);
  }
  node* get_path_from_root(int i) {
    node* v = nodes[i];
    expose(v);
    return v;
  }
  template <typename... T>
  void apply(int i, T... args) {
    node* v = nodes[i];
    splay_tree::apply(v, args...);
  }
};
int n, m, q;
long long C;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(15);
  cin >> n >> m >> q >> C;
  vector<int> f(n);
  for (int i = 0; i < n; i++) {
    cin >> f[i];
    f[i]--;
  }
  tree g(n);
  for (int i = 1; i < n; i++) {
    int u;
    cin >> u;
    u--;
    g.add(u, i);
  }
  vector<int> c(m);
  for (int i = 0; i < m; i++) {
    cin >> c[i];
  }
  g.prep();
  vector<vector<int>> id(m, vector<int>(1, 0));
  for (int i = 0; i < n; i++) id[f[i]].push_back(i);
  vector<vector<int>> que;
  for (int i = 0; i < q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int x, w;
      cin >> x >> w;
      w--;
      x--;
      id[w].push_back(x);
      que.push_back({t, x, w});
    } else {
      int w;
      cin >> w;
      w--;
      que.push_back({t, w});
    }
  }
  vector<link_cut_tree<true>> cur(m);
  vector<map<int, int>> to(m);
  vector<long long> res(m);
  vector<long long> total(m);
  vector<vector<int>> bbug(m);
  for (int i = 0; i < m; i++) {
    auto foo = g.to_tree(id[i]);
    vector<int> par = foo.second;
    bbug[i] = par;
    to[i] = foo.first;
    int sz = par.size();
    cur[i] = link_cut_tree<true>(sz);
    vector<int> ps(sz);
    for (auto it : to[i]) ps[it.second] = it.first;
    vector<int> cnt(sz);
    for (int j = sz - 1; j > -1; j--) {
      cnt[j] += f[ps[j]] == i;
      int mu = g.dep[ps[j]] - (par[j] == -1 ? -1 : g.dep[ps[par[j]]]);
      res[i] += 1LL * cnt[j] * cnt[j] * mu;
      total[i] += 1LL * cnt[j] * mu;
      if (~par[j]) cnt[par[j]] += cnt[j];
      cur[i].nodes[j]->dep = mu;
      cur[i].nodes[j]->total_dep = mu;
      cur[i].nodes[j]->sum = 1LL * cnt[j] * mu;
      cur[i].nodes[j]->total_sum = 1LL * cnt[j] * mu;
    }
    for (int j = sz - 1; j > 0; j--) {
      cur[i].link(j, par[j]);
    }
  }
  for (auto q : que) {
    int t = q[0];
    if (t == 1) {
      int x = q[1], w = q[2];
      res[f[x]] += g.dep[x] + 1;
      total[f[x]] -= g.dep[x] + 1;
      int i = f[x];
      node* v = cur[i].get_path_from_root(to[i][x]);
      assert(g.dep[x] + 1 == v->total_dep);
      res[i] -= v->total_sum * 2;
      cur[i].apply(v->id, -1);
      f[x] = w;
      res[f[x]] += g.dep[x] + 1;
      total[f[x]] += g.dep[x] + 1;
      node* u = cur[w].get_path_from_root(to[w][x]);
      assert(g.dep[x] + 1 == u->total_dep);
      res[w] += u->total_sum * 2;
      cur[w].apply(u->id, 1);
    } else {
      int w = q[1];
      double ans = res[w] * c[w] * c[w] - 2LL * c[w] * total[w] * C + n * C * C;
      ans /= n;
      cout << ans << "\n";
    }
  }
  return 0;
}